Isosceles right triangle problem (solids with know cross-sections)

I have a isosceles right triangle with a known base (square root of (4-x^2) how can I find the height? Do I need to use a law of sin/cos or is there an easier way?

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- Feb 3rd 2010, 12:00 PMicu812Isosceles right triangle problem
Isosceles right triangle problem (solids with know cross-sections)

I have a isosceles right triangle with a known base (square root of (4-x^2) how can I find the height? Do I need to use a law of sin/cos or is there an easier way? - Feb 3rd 2010, 01:00 PMCalculus26
The height is the length of one of the equal sides the area is 1/2s^2

since it is a rt triangle s^2 +s^2 = h^2

s= h/sqrt(2)

in your eg h = 4-x^2 - Feb 3rd 2010, 01:02 PMSoroban
Hello, icu812!

Did you make a sketch?

Quote:

I have a isosceles right triangle with a known base:\ .$\displaystyle b \,=\,\sqrt{4-x^2}$

How can I find the height?

Code:`*`

* | *

* | *

* |b/2 *

* | *

* 45° | 45° *

* * * * * * * * *

: - b/2 - - : - - b/2 - :

: - - - - - b - - - - - :

Got it?

- Feb 3rd 2010, 01:12 PMicu812
Yes thanks a lot. It has been probaly 15 years since I had geometry. We just did solids of revolution, and started solids with known cross-section today. (Itwasntme)